# Prove that $n\leq a_1+a_2+…+a_n \leq n+1$

If $$a_1=2,a_{n+1}=\sqrt{a_n+8}-\sqrt{a_n+3}$$. Prove that $$n\leq a_1+a_2+...+a_n \leq n+1$$ for every $$n\ge1$$ and $$\lim a_n=1$$.

I have showed that by squaring and inequality techniques:

1. $$a_i<\sqrt{3}$$ for every $$i>1$$.
2. If $$a_i>1$$ then $$a_{i+1}<1$$ for every $$i\ge 1$$

I think that $$\sqrt{3}$$ can be improved, but I am not sure if it's useful.

We have

$$a_{n+1}-1 = (a_n-1) \left( \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right)$$

and

$$(a_{n+1}-1)+(a_n-1) = (a_n-1) \left( 1+ \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right)$$,

a) $$\left| \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right| < \left| \cfrac{1}{3+\sqrt{8+a_n}} + \cfrac{1}{2+\sqrt{3+a_n}} \right| < \left| \cfrac{1}{3+\sqrt{8+0}} + \cfrac{1}{2+\sqrt{3+0}} \right| = 5-(\sqrt 3 + \sqrt8) < 1$$

This way we have $$|a_{n+1}-1| < q |a_n-1| < q^n |a_1-1|$$ by induction. This way, $$|a_{n+1}-1|$$ tends to 0.

b) We have to show some results:

• $$\left( \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right) < 0$$, therefore $$a_{n+1}-1$$ and $$a_n-1$$ have opposite signals. But $$a_1-1=1>0$$, then the even guys are negative, the odd ones are positive: $$a_{2k} < 1 < a_{2k+1}$$.

• $$\left( 1+ \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right) > 0$$, therefore $$a_{n+1}+a_n-2$$ and $$a_n-1$$ have the same signals: $$a_{2k}+a_{2k+1} < 2 < a_{2k+1}+a_{2k+2}$$

Now we have two cases to consider:

• $$(a_1+a_2)+\ldots+(a_{2k+1}+a_{2k+2}) > 2+\ldots+2=2k+2$$ and $$a_1+(a_2+a_3)+\ldots+(a_{2k}+a_{2k+1})+a_{2k+2} < 2+\ldots+2+1=2k+3$$

• $$(a_1+a_2)+\ldots+(a_{2k+1}) > 2+\ldots+1=2k+1$$ and $$a_1+(a_2+a_3)+\ldots+(a_{2k}+a_{2k+1}) < 2+\ldots+2+2=2k+2$$